Inverse proof

Prove that if f:X -> Y is a discontinuous bijection then f-1:Y -> X is also discontinuous.

2. Relevant equations

N/A

3. The attempt at a solution

The contrapositive of this statement is that if f-1:Y -> X is continuous then f:X -> Y is continuous. Since f is bijective its invertable; hence, all I need to prove is, if f:X -> Y is continuous then f-1:Y -> X is continuous.

If f is continuous there exits a value delta for every epsilon such that delta = g*epsilon, where g is a function of epsilon.

We use the epislon delta method to prove f-1 is continuous. Hence, for all epsilon greater than zero there exists a delta greater than zero such that 0 < abs(y - k) < delta and 0 < abs(f-1(y) - f-1(k)) < epsilon. Hence, we may rewrite this as 0 < abs(f(x) - f(c)) < delta and 0 < abs(x - c) < epsilon. Since, this is merely the situation for f with epsilon and delta switched it follows that delta = epsilon/g. Therefore f-1:Y -> X is continuous. Q.E.D.

I think some of my reasoning isn't particularly great, so any suggestions on how to correct this proof are welcome. Thanks.

The contrapositive of this statement is that if f-1:Y -> X is continuous then f:X -> Y is continuous. Since f is bijective its invertable; hence, all I need to prove is, if f:X -> Y is continuous then f-1:Y -> X is continuous.

You can't prove an implication by proving its converse. While that statement is also true, it does not show that "if f-1:Y -> X is continuous then f:X -> Y is continuous". If you're trying to prove the contrapositive statement the straightforward constructive way, you have to start by assuming that f-1 is continuous.

Sorry. I figured that since if a function is completely invertable then it's inverse is completely invertable as well, and since any function f that has a complete inverse g can also be called g-1, it followed that if f is a continuous bijection then f-1 is also continuous or equivalently stated if g-1 is a continuous bijection, then g is also continuous. I should have caught that and proved it probably. It should be simple to fix that part though.

If f-1 is continuous then for every epsilon (e) there exists a delta (d) given by d = g*e where g is a function of e.

We use the epsilon delta method to prove f is continuous. For all e > 0 there exists a d > 0 such that 0 < abs(x - c) < d and 0 < abs(f(x) - f(c)) < e. Which is equivalent to 0 < abs(f-1(y) - f-1(k)) < d and 0 < abs(y - k) < e. Since this is merely the situation for f-1 with e and d switched it follows that d = e/g. Q.E.D.

Prove that if f:X -> Y is a discontinuous bijection then f-1:Y -> X is also discontinuous.

3. The attempt at a solution

The contrapositive of this statement is that if f-1:Y -> X is continuous then f:X -> Y is continuous. Since f is bijective its invertable; hence, all I need to prove is, if f:X -> Y is continuous then f-1:Y -> X is continuous.

If f is continuous there exits a value delta for every epsilon such that delta = g*epsilon, where g is a function of epsilon.I can't make any sense of this.

We use the epislon delta method to prove f-1 is continuous. Hence, for all epsilon greater than zero there exists a delta greater than zero such that 0 < abs(y - k) < delta and 0 < abs(f-1(y) - f-1(k)) < epsilon. Hence, we may rewrite this as 0 < abs(f(x) - f(c)) < delta and 0 < abs(x - c) < epsilon. Since, this is merely the situation for f with epsilon and delta switched it follows that delta = epsilon/g. Therefore f-1:Y -> X is continuous. Q.E.D. No: think about what you want to show. You want to show that, for each c in X, if I give you a B>0 then you can find an A>0 such that |x-c|<A implies that |f(x)-f(c)|<B. Your 'proof' can only handle the case when the B I give you is the delta you found, which depended on your original epsilon!

Without certain conditions on X the statement to prove is false, anyway. A common counterexample is with X = [0,1) U (1,2], f(x)=x on [0,1] and f(x)=x+1 on (1,2].
EDIT: X should be [0,2].

I'll go along with your other criticisms though. I'll try to explain my reasoning more clearly. First, what I mean by d = g*e: Suppose d = 2e +1, this can be expressed in the form d = g*e where g = 2 + 1/e, I'm simply trying to generalize something. Given 0 < abs(x - c) < e and 0 < abs(f(x) - f(c)) < d. If we go about the same process used to find e we would get e = g*d where g is a function of delta. Now I suppose my "proof" only works if you can explicitly express that in terms of d which is what I presume you were pointing out.